Re: [微積] 一題Jacobian變換
※ 引述《james1201 ()》之銘言：
: 2
: 2 u (v-u+1)
: Evaluate the iterated integral S S e dvdu.
: 1 2u-2
: 這題我是用Jacobian變換去解 令x=v-u+1來嘗試
: y=v
: 可是弄了老半天好像會卡住
: 請問有更好的解法嗎？
用另外的方法可以求, 不過應該不是出這題的本意
u
寫 f(u) = ∫exp[(v-u+1)^2]dv, 則
2u-2
d d u ∂
f'(u) = exp[(u-u+1)^2]--u - exp[(2u-2-u+1)^2]--(2u-2) + ∫----exp[(v-u+1)^2]dv
du du 2u-2 ∂u
u
= e - 2exp[(u-1)^2] + ∫(-2)(v-u+1)exp[(v-u+1)^2]dv
2u-2
|v = u
= e - 2exp[(u-1)^2] - exp[(v-u+1)^2]|
|v = 2u-2
= -exp[(u-1)^2]
2 |2 2
=> 所求 = ∫f(u)du = uf(u)| - ∫uf'(u)du
1 |1 1
1 2
= -∫exp(v^2)dv + ∫u exp[(u-1)^2] du
0 1
1
= ∫[ (t+1)exp(t^2) - exp(t^2) ]dt (u=t+1, v=t)
0
1 |1 e - 1
= ---exp(t^2)| = -------
2 |0 2
--
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◆ From: 59.115.145.87
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